What is the integral from 0 to 1 of t^3(4+t^4)^3 ?

The integral from 0 to 1 of t^3(4+t^4)^3 is 369/16

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integral from 0 to 1 of t^3(4+t^4)^3

\int_{0}^{1} t^{3} (4 + t^{4})^{3} d t

\frac{369}{16}

Decimal Notation

23.0625

Solution steps

\int_{0}^{1} t^{3} (4 + t^{4})^{3} d t

Expand

t^{3} (4 + t^{4})^{3} : 64 t^{3} + 48 t^{7} + 12 t^{11} + t^{15}

= \int_{0}^{1} 64 t^{3} + 48 t^{7} + 12 t^{11} + t^{15} d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{1} 64 t^{3} d t + \int_{0}^{1} 48 t^{7} d t + \int_{0}^{1} 12 t^{11} d t + \int_{0}^{1} t^{15} d t

\int_{0}^{1} 64 t^{3} d t = 16

\int_{0}^{1} 48 t^{7} d t = 6

\int_{0}^{1} 12 t^{11} d t = 1

\int_{0}^{1} t^{15} d t = \frac{1}{16}

= 16 + 6 + 1 + \frac{1}{16}

Simplify

16 + 6 + 1 + \frac{1}{16} : \frac{369}{16}

= \frac{369}{16}

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What is the integral from 0 to 1 of t^3(4+t^4)^3 ?
The integral from 0 to 1 of t^3(4+t^4)^3 is 369/16